Properties of functions provided that the integral equation $\int_x^1 a(y-x)a(y) \, dy = \int_x^1 b(y-x)b(y) \, dy$ holds for $x \in [0,1]$

Let $$a,b : [0,1] \to \mathbb R$$ are two functions (e.g. suppose that they are in $$L^2[0,1]$$ or are $$N$$-times continuously differentiable). Now suppose that $$\int_x^1 a(y-x)a(y) \, dy = \int_x^1 b(y-x)b(y) \, dy$$ for every $$x \in [0,1]$$. I'm interested in properties of $$a,b$$ and how they are related, provided that $$a,b$$ satisfy the above equation of integrals for every $$x \in [0,1]$$. Clearly, they do not have to be equal since if $$a=-b$$ then the above equation holds. I'm wondering of this is the only thing which can happen. Did someone of you came across such a problem before?

There are plenty of examples of such pairs of functions. First, extend your $$a$$ and $$b$$ to the whole real line by setting them to $$0$$ outside the interval $$[0,1]$$. So your relation becomes $$\int_{-\infty}^\infty a(y-x)a(y)dy=\int_{-\infty}^\infty b(y-x)b(y)dy.$$ Your condition is that this must hold only for $$x\in[0,1]$$, but I will construct many examples in which this will hold for all real $$x$$. Suppose first that this equation holds for all real $$x$$. Now let $$f(x)=a(-x),\; g(x)=g(-x)$$. Then your equation is equivalent to $$a\star f=b\star g,$$ where $$*$$ is the convolution. Taking Fourier transforms and using $$\hat{f}=\hat{a}(-s)$$, and similar for $$\hat{g}$$, we obtain $$\hat{f}(s)\hat{f}(-s)=\hat{g}(s)\hat{g}(-s),\quad\quad\quad\quad\quad\quad (1)$$ and this is equivalent to the assumption that your equation holds for all real $$x$$. Now $$\hat{f},\hat{g}$$ are entire functions of exponential type, bounded in the lower half-plane (by the Wiener-Paley theorem), so $$\hat{f}(s)=e^{cs}\prod_j\left(1-\frac{s}{s_j}\right)e^{s/s_j},$$ and they are determined by their zeros and the exponential factor in front. In equation (1), the exponential factors cancel, so you already have plenty of examples: take $$a(x)$$ with some small support on $$(0,1)$$ and let $$b$$ be a small shift: $$b(x)=a(x-\epsilon)$$ so that it also has support on $$(0,1)$$.

But there is much more. Let $$Z$$ be the set of zeros of $$\hat{f}$$. Then zeros of $$\hat{f}(s)\hat{f}(-s)$$ are $$Z\cup(-Z)$$, and by decomposing this union in some other way: $$Z\cup(-Z)=Z_1\cup(-Z_1)$$ you obtain a function $$\hat{g}$$ for which (1) holds. Typically $$\hat{f}$$ will have infinitely many zeros, so such decomposition can be done in infinitely many ways, by redistributing, say, finitely many of zeros.

To make sure that all these $$\hat{f},\hat{g}$$ are really Fourier transforms of some functions supported on $$[-1,0]$$ one uses the Wiener-Paley theorem.

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I have seen something similar. But not the same. The set of continuous real-valued functions $$C[0,\infty)$$ with pointwise addition and convolution for multiplication $$(f * g)(x) = \int_0^x f(x-t)g(t)\;dt$$ is an integral domain (without unit). Consequently, such a function can have at most $$2$$ square-roots... $$f*f = g*g \\ f*f-g*g = 0 \\ (f-g)*(f+g)=0 \\ f-g = 0 \text{ or } f+g = 0 \\ f=g \text{ or } f=-g$$

That "integral domain" result is due to Titchmarsh .

For the OP question, a similar approach would be to let $$c=a-b, d=a+b$$ and then ask: is it true that $$\int_x^1 c(y-x)d(y)\;dy = 0\quad\text{for all } x \in [0,1] \\ \Downarrow\\ \big[c(x) = 0 \quad\text{for all } x \in [0,1]\big]\quad\text{ or }\quad \big[d(x) = 0 \quad\text{for all } x \in [0,1]\big]$$